Question Number 192470 by Spillover last updated on 19/May/23
Answered by Spillover last updated on 19/May/23
![∫_0 ^(√2) (√(1+(2x)^2 )) dx Let 2x=sinh θ dx= ((cosh θdθ)/2) ∫_0 ^(√2) (√(1+(2x)^2 )) dx ∫_0 ^(√2) (√(1+sinh^2 θ)) . ((cosh θdθ)/2)dx (1/2)∫cosh^2 θdθ (1/2)∫(((1+cosh 2θ)/2)) (1/4)∫dθ+(1/4)∫cosh 2θ =(1/4)θ+(1/8)sinh2θ (1/4)θ+(1/8)sinh2θ=(1/4)θ+(1/4)sinhθcosh θ 2x=sinh θ θ=sinh^(−1) 2x coshθ=(√(1+sinh^2 θ)) (1/4)sinh^(−1) 2x+((2x)/4)(√(1+4x^2 )) (1/4)ln ∣2x+(√(1+4x^2 ))∣+(x/2)(√(1+4x^2 )) [(1/4)ln ∣2x+(√(1+4x^2 ))∣+(x/2)(√(1+4x^2 )) ]_0 ^(√2) ((3(√2))/2)+(1/4)ln ∣3+2(√(3 ))∣−0 ((3(√2))/2)+(1/4)ln ∣3+2(√(2 ))∣](Q192529.png)
$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \sqrt{\mathrm{1}+\left(\mathrm{2}{x}\right)^{\mathrm{2}} }\:{dx} \\ $$$${Let}\:\:\:\mathrm{2}{x}=\mathrm{sinh}\:\theta\:\:\:\:\:\:\:\:\:\:\:{dx}=\:\frac{\mathrm{cosh}\:\theta{d}\theta}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \sqrt{\mathrm{1}+\left(\mathrm{2}{x}\right)^{\mathrm{2}} }\:{dx}\:\:\:\:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \sqrt{\mathrm{1}+\mathrm{sinh}\:^{\mathrm{2}} \theta}\:.\:\frac{\mathrm{cosh}\:\theta{d}\theta}{\mathrm{2}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{cosh}\:^{\mathrm{2}} \theta{d}\theta\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}+\mathrm{cosh}\:\mathrm{2}\theta}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int{d}\theta+\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{cosh}\:\mathrm{2}\theta\:\:=\frac{\mathrm{1}}{\mathrm{4}}\theta+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sinh2}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\theta+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sinh2}\theta=\frac{\mathrm{1}}{\mathrm{4}}\theta+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sinh}\theta\mathrm{cosh}\:\theta \\ $$$$\:\mathrm{2}{x}=\mathrm{sinh}\:\theta\:\:\:\:\:\:\theta=\mathrm{sinh}\:^{−\mathrm{1}} \mathrm{2}{x}\:\:\:\:\:\mathrm{cosh}\theta=\sqrt{\mathrm{1}+\mathrm{sinh}\:^{\mathrm{2}} \theta}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sinh}\:^{−\mathrm{1}} \mathrm{2}{x}+\frac{\mathrm{2}{x}}{\mathrm{4}}\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\mid+\frac{{x}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\: \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\mid+\frac{{x}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\:\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}\:}\mid−\mathrm{0} \\ $$$$\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}\:}\mid\: \\ $$